Which is larger $(101!)^{100}$ or $(100!)^{101}$ [duplicate]

I am supposed to tell which one of $$(101!)^{100}$$ and $$(100!)^{101}$$ is larger. I am trying to use the behavior of the function $$f(x)=x^{1/x}$$ as is a standard technique to dealing with questions of this sort. Here is what I have so far.

\begin{aligned}(101!)^{100!}&\lt (100!)^{101!}\\ (101!)^{100} &\lt (100!)^{101\times 100}\end{aligned}

Any ideas on how to proceed. Thanks.

• Do whatever cancellations you can and you will get the answer easily. Oct 22, 2020 at 7:24
• There's a factorial in the power as well? Oct 22, 2020 at 7:25
• @Gray_Rhino No, there isn't. Oct 22, 2020 at 7:28
• @ParasKhosla But in the body of your post, you've used factorial in the power Oct 22, 2020 at 7:36
• This is a duplicate of this question, where $n = 100$, and raising both sides by $100 \cdot 101$ gives you this inequality. I am linking to the abstract duplicate Inequality from Chapter 5 of the book *How to Think Like a Mathematician* because it is much more well-known. Oct 22, 2020 at 7:56

Well

$$(101!)^{100}\cdot (101!)=\color{green}{(101!)^{101}}$$ while: $$(100!)^{101}\cdot (101)^{101}=\color{green}{(101!)^{101}}$$

the one you have to multiply by the larger number is smaller

Simplify: \begin{align} \frac{(101!)^{100}}{(100!)^{101}}&=\frac{(100!\times 101)^{100}}{(100!)^{100}\times(100!)}\\ &=\frac{(100!)^{100}\times 101^{100}}{(100!)^{100}\times(100!)}\\ &=\frac{101^{100}}{100!} \end{align} Now, $$\frac{101^{100}}{100!}=\frac{101\times101\times101\times\dots_{100\mathrm{~times}}}{100\times99\times98\times\dots2\times1}=\frac{101}{100}\times\frac{101}{99}\times\frac{101}{98}\times\dots\frac{101}{2}\times\frac{101}{1}>1.$$ You get

$$\frac{(101!)^{100}}{(100!)^{101}}=\frac{101^{100}}{100!}>1\Rightarrow(101!)^{100}>(100!)^{101}$$

Consider $$A_n=\big[(n+1)!\big]^n \qquad \text{and}\qquad B_n=\big[n!\big]^{n+1}$$ $$R_n=\frac{A_n}{B_n}=\frac{\big[(n+1)!\big]^n } {\big[n!\big]^{n+1} }$$ Take logarithms $$\log(R_n)=n \log ((n+1)!)-(n+1) \log (n!)$$ Use Stirling approximation and continue with Taylor expansions $$\log(R_n)=n+\left(1-\frac{1}{2} \log (2 \pi n)\right)-\frac{7}{12 n}+O\left(\frac{1}{n^2}\right)$$ $$R_n\sim\frac{e^{n+1}}{\sqrt{2 \pi n}}$$

For $$n=100$$, the above would give $$R_{100}\sim 2.92\times 10^{42}$$ while the exact value would be $$2.90\times 10^{42}$$